It is not difficult to prove that $0/0$ is not equal to any constant $c \ | \ c/c = 1 \ \land \ c \in \mathbb{R} \cup \mathbb{C}$.
First assume that $0/0 = c$. Consider the true equation $2\times0=0$. Based on our initial assumption, dividing the equation by zero gives $2\times0/0=0/0 \ \Rightarrow \ 2c=c$. Based on our definition of $c$, dividing the equation by $c$ gives $2=1$. However, $2 \ne 1$. $\ \therefore\ $ $0/0 \ne c.\quad\blacksquare$
Sorry for the incomprehensible spacing. Anyways, whenever I attempt to prove that $0/0 \ne 0$, I always end up arriving at $0=0$. Can someone find (or make) a proof? I couldn't find one online either.
Just some clarifications: In in the argument that ends with a box, I am attempting to prove that $0/0 \ne 0$.
Extra clarifications: This is not a duplicate. I am not asking for an intuitive proof that division by zero is nonsensical. I am asking for a concrete mathematical proof that $0/0 \ne 0$. And sorry about my previous edit, that was incorrect but I fixed it.
The quotient $a/b$ is the unique number $c$ such that $bc=a.$
So the quotient $0/0$ should be the unique number $c$ such that $0c=0.$
The problem is non-uniqueness.
If one considers $5/0$ rather than $0/0$ then the problem is non-existence of any number $c$ for
which $0c=5.$
Either non-existence of non-uniqueness makes division by $0$ undefined.